Problem 9 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{1}{z} \frac{d z}{d t}=5, \quad z(1)=5$$

Step-by-Step Solution

Answer

The solution is \( z(t) = 5e^{5t-5} \).

Step 1: Separate the Variables

Begin by rewriting the given differential equation \( \frac{1}{z} \frac{dz}{dt} = 5 \). The goal is to separate \( z \) and \( t \) on opposite sides. Multiply both sides by \( z \): \[ \frac{dz}{dt} = 5z \]

Step 2: Integrate Both Sides

After separating the variables, integrate both sides. The left side with respect to \( z \) and the right side with respect to \( t \): \[ \int \frac{1}{z} \, dz = 5 \int 1 \, dt \]

Step 3: Solve the Integrals

Calculate the integrals found in the previous step: \[ \ln |z| = 5t + C \] where \( C \) is the constant of integration.

Step 4: Solve for the Constant Using Initial Condition

Use the initial condition \( z(1)=5 \) to find \( C \). Substituting into the integrated result, we have: \( \ln |5| = 5(1) + C \). Therefore, \[ C = \ln 5 - 5 \]

Step 5: Express the Solution Explicitly

Substitute the value of \( C \) back into the equation. Therefore, \( \ln |z| = 5t + \ln 5 - 5 \). Exponentiate both sides to solve for \( z \): \[ z = e^{5t + \ln 5 - 5} = 5e^{5t-5} \] using the property \( e^{\ln a} = a \).

Key Concepts

Separation of Variables

Separation of variables is a method used to solve differential equations by dividing them into parts where each variable is isolated on different sides of the equation. This approach is particularly useful for equations that can be rearranged in such a way. Consider the equation: \( \frac{1}{z} \frac{dz}{dt} = 5 \). The goal is to isolate \( z \) and \( t \) so we can deal with them individually:

• Multiply both sides by \( z \) to get \( \frac{dz}{dt} = 5z \).
• This allows us to express the change in \( z \) relative to \( t \) alone.

This separation is the first critical step before proceeding to integration, making it easier to find a solution that includes both variables.

Initial Conditions

Initial conditions are specific values that a solution to a differential equation must satisfy. These conditions provide a specific context for the solution, helping us determine exact values for any constants that arise during the integration process. Consider the equation after separation:

• Without an initial condition, the solution could be any function that fits the form of the general solution.
• Using \( z(1) = 5 \), we apply this condition after integrating to find any constants, like \( C \), to ensure the solution meets the initial criteria.

This step tailors our general answer to a specific situation described by the differential equation.

Integration

Integration is a crucial process in solving differential equations, especially after the separation of variables. It involves finding the antiderivative or integral of both sides of an equation. After separating our variables, we set up integrations:

• \( \int \frac{1}{z} \, dz = 5 \int 1 \, dt \)
• Solving these integrals involves recognizing standard forms and applying techniques to solve them.

For our exercise:

• \( \ln |z| = 5t + C \) is derived, where \( C \) represents an arbitrary constant.

Integration brings us closer to expressing a complete solution, bridging the unknowns through calculus.

Solutions to Differential Equations

When solving differential equations, the end goal is to express the dependent variable in terms of the independent variable. Initially separated and integrated, our solutions form a pattern that follows:

• Taking the exponential of both sides helps clear logarithms: \( z = e^{5t + C} \).
• By using our initial conditions, we adapted this further to \( z = 5e^{5t-5} \).
• This final form presents \( z \) analytically in terms of \( t \), solving the original differential equation entirely.

These steps allow us to transform a complex differential equation into a practically applicable solution, significant in many fields needing precise calculations.